Yes, you are right, they don't comute in general. I confused it with another situation, I'm sorry.I think it's wrong in general since ##\hat{C}(t,t_0)## and ##\hat{H}(t)## do not commute in general.
I have already read it in a book, but I didn't understand what it means. The most books say that all operators are time-independent in this picture. So, how can they have an explicit dependence on time?In the Schrödinger picture all time dependence of operators that represent observables are by definition explict time dependence.
To be more specific, use as example a non-relativistic single particle. Then the fundamental set of operators are the position, momentum, and spin operators ##\hat{\vec{x}}##, ##\hat{\vec{p}}##, and ##\hat{\vec{\sigma}}##, all of which by definition do not explicitly depend on time.Yes, you are right, they don't comute in general. I confused it with another situation, I'm sorry.
I have already read it in a book, but I didn't understand what it means. The most books say that all operators are time-independent in this picture. So, how can they have an explicit dependence on time?
I'm sorry for insisting on this question, I am really confused about pictures in quantum mechanics.